#### To determine

**To estimate:**

The amount of water that flowed into Lake Lanier from July *18th, 2013*, at *7:30 AM* to July *26th*, at *7:30 AM*

#### Answer

*2,954,707,200 ft3*

#### Explanation

**1) Concept:**

Use the formula for the amount of water that flowed into Lake Lanier and calculate it by using the midpoint rule.

**2) Formula:**

The amount of water flowed is *∫t1t2r(t)dt* where *r(t)* is the inflow rate during the time *t1* to *t2*.

**3) Calculation:**

Let *r(t)* be the in flow rate which is given in *ft3/s*.

The amount of water that flowed into Lake Lanier from July *18th, 2013*, at *7:30 AM* to July *26th*, at *7:30 AM*, that is, in *8* days is given by

*∫08r(t)dt*

Now, find *∫08r(t)dt* by using the midpoint rule,

With *n=4*, *a=0* and *b=8* we have the width of interval is

∆t=b-an

Substitute the values;

∆t=8-04

∆t=2

And the midpoints are

t1-=12t0+t1=120+2=122=1

t2-=12t1+t2=122+4=126=3

t3-=12t2+t3=124+6=1210=5

t4-=12t3+t4=126+8=1214=7

So, an estimate for the value of the integral is

∫04r(t)dt≈M4=∑i=14r(ti-)∆t

=∆x(rt1-+rt2-+rt3-+rt4-)

=∆x(r1+r3+r5+r(7))

From the given table,

r1=6401

r3=4249

r5=3821

r7=2628

Substitute these values in Riemann sum.

M4=2(6401+4249+3821+2628)

Simplify,

M4=217,099=34,198

Now, in one day, there are *24·602* seconds.

Multiply by these seconds.

Therefore, the amount of water that flowed into Lake Lanier is *34,198·24·602 ft3*

*=2,954,707,200 ft3*

**Conclusion:**

The amount of water that flowed into Lake Lanier from *7:30 AM* on July *18th, 2013* to *7:30 AM* on July *26th* is *2,954,707,200 ft3*.